Is Integral is considered defined even when one method gives defined results and other undefined results

In the $\frac{0}{0}$ case, we can apply L'Hôpital's rule to obtain $$lim_{t\to 0}\frac{e^{tB}-e^{tA}}{t}=lim_{t\to 0}\frac{Be^{tB}-Ae^{tA}}{1}=B-A$$ so the two results agree. If we integral a continuous function of two variables with respect to one variable, the result should be a continuous function of the other.

In general, if $F(t)=\int_a^b f(t,x)dx$, we would define, for fixed $t_0$, $F(t_0)=\int_a^b f(t_0,x)dx$ rather than evaluating for general $t$ and then setting $t=t_0$. But assuming $f$ is sufficiently nice, them the two will agree so long as we take limits

You are using the formula $\int e^{tx}\, dx = \frac{e^{tx}}{t} + C$, but this rule is not valid when $t=0$. You should actually use

$$\int e^{tx}\, dx = \begin{cases} \frac{e^{tx}}{t} + C &\text{ if } t \neq 0 \\ x + C &\text{ if } t = 0.\\ \end{cases}$$

Then the two results are the same. In general, when you use a formula involving a division be aware that the formula will not be valid when the denominator is $0$. This is often an unstated assumption, so you should check if both sides a given equation are undefined or not in that case.